Law of conservation of energy states that the energy can neither be created nor destroyed but can be transformed from one form to another.
Let us now prove that the above law holds good in the case of a freely falling body.
Let a body of mass 'm' placed at a height 'h' above the ground, start falling down from rest.
In this case we have to show that the total energy (potential energy + kinetic energy) of the body A, B and C remains constant i.e, potential energy is completely transformed into kinetic energy.
Body of mass m placed at a height h
At A,
Potential energy = mgh
Kinetic energy = 0 [the velocity is zero as the object is initially at
rest]
Total energy at A = Potential energy + Kinetic energy.
Total energy at A = mgh …1
At B,
Potential energy = mgh
= mg(h - x) [Height from the ground is (h-x)]
Potential energy = mgh - mgx
Kinetic energy = ½ mv2
The body covers the distance x with a velocity v. We make use of the third equation of motion to obtain velocity of the body.
Here, u=0, a=g and s=x
Kinetic energy = mgx
Total energy at B = Potential energy + Kinetic energy
Total energy at B = mgh …2
At C,
Potential energy = m x g x 0
Potential energy = 0
Kinetic energy = ½ mv2
The freely falling body has covered the distance h.
Here, u=0, a=g and s=h
Kinetic energy = ½ mv2
Kinetic energy = mgh
Total energy at C = Potential energy + Kinetic energy
Total energy at C = mgh …3
It is clear from equations 1, 2 and 3 that the total energy of the body remains constant at every point. Thus, we conclude that law of conservation of energy holds good in the case of a freely falling body.
Test
Gaurav Seth 7 years, 3 months ago
Law of conservation of energy states that the energy can neither be created nor destroyed but can be transformed from one form to another.
Let us now prove that the above law holds good in the case of a freely falling body.
Let a body of mass 'm' placed at a height 'h' above the ground, start falling down from rest.
In this case we have to show that the total energy (potential energy + kinetic energy) of the body A, B and C remains constant i.e, potential energy is completely transformed into kinetic energy.
Body of mass m placed at a height h
At A,
Potential energy = mgh
Kinetic energy = 0 [the velocity is zero as the object is initially at
rest]
Total energy at A = mgh …1
At B,
Potential energy = mgh
= mg(h - x) [Height from the ground is (h-x)]
Potential energy = mgh - mgx
Kinetic energy = ½ mv2
The body covers the distance x with a velocity v. We make use of the third equation of motion to obtain velocity of the body.
Here, u=0, a=g and s=x
Kinetic energy = mgx
Total energy at B = mgh …2
At C,
Potential energy = m x g x 0
Potential energy = 0
Kinetic energy = ½ mv2
The freely falling body has covered the distance h.
Here, u=0, a=g and s=h
Kinetic energy = ½ mv2
Kinetic energy = mgh
Total energy at C = mgh …3
It is clear from equations 1, 2 and 3 that the total energy of the body remains constant at every point. Thus, we conclude that law of conservation of energy holds good in the case of a freely falling body.
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